1. Why do plants need both mitochondria and chloroplasts?
Plants need both chloroplasts and mitochondria because plants produce convert solar energy into chemical energy but they also need energy for growth and metabolic processes. The conversion of solar energy into chemical energy in the form of glucose occurs during photosynthesis, which occurs in the chloroplasts. Plants, just like all other living organisms, need energy. This energy is produced during cellular respiration, which takes place in the mitochondria.
2. You have a leaf from each of two very different plants. One leaf has more pigments than the other. Which leaf would have the greater photosynthetic rate, assuming all affecting factors are equal.
The leaf with more pigments will have a greater photosynthetic rate, because more pigments allow more photosynthesis to occur, increasing the rate of photosynthesis.
3. Outline the effect of temperature, light intensity, and carbon dioxide concentration on the rate of photosynthesis.
As temperature increases, the rate of photosynthesis increases more and more steeply until it reaches an optimum temperature. Above the optimum temperature, the rate of photosynthesis falls steeply. At low to medium light intensities, the rate of photosynthesis is dirtily proportional to light intensity. At high light intensities, the rate of photosynthesis plateaus. At very low carbon dioxide concentrations, no photosynthesis occurs. At low to fairly high carbon dioxide concentrations, the rate of photosynthesis is positively correlated with CO2 concentration. At very high CO2 concentrations, the rate of photosynthesis plateaus.
Monday, October 28, 2013
Monday, October 14, 2013
Warm Up 10/14
1. Describe chemiosmosis as it relates to oxidative phosphorylation.
Chemiosmosis occurs at the inner membrane. It involves the movement of protons (hydrogen ions) across a membrane (down its concentration gradient) to provide energy so that oxidative phosphorylation (ATP synthesis) can occur. Hydrogen ions build up in the intermembrane space, creating a high concentration and a proton force. This allows the movement of hydrogen ions through the enzyme ATP synthase. The enzyme uses energy from the hydrogen flow to couple phosphate with ADP to produce ATP.
2. What is the role of oxygen in the ETC?
Oxygen is the final electron acceptor in the ETC. Oxygen also accepts hydrogen ions to form water. If oxygen is not available, the electron flow along the ETC stops, but glycolysis can still occur.
Chemiosmosis occurs at the inner membrane. It involves the movement of protons (hydrogen ions) across a membrane (down its concentration gradient) to provide energy so that oxidative phosphorylation (ATP synthesis) can occur. Hydrogen ions build up in the intermembrane space, creating a high concentration and a proton force. This allows the movement of hydrogen ions through the enzyme ATP synthase. The enzyme uses energy from the hydrogen flow to couple phosphate with ADP to produce ATP.
2. What is the role of oxygen in the ETC?
Oxygen is the final electron acceptor in the ETC. Oxygen also accepts hydrogen ions to form water. If oxygen is not available, the electron flow along the ETC stops, but glycolysis can still occur.
Thursday, October 10, 2013
Warm Up 10/10
1. What is denaturation?
Denaturation is changing the structure of an enzyme (or other protein) so that it can no longer carry out its function. Enzymes are denatured at very high temperatures and extreme pHs.
2.What does a graph for increasing substrate concentration of an enzyme look like? Describe it...
At low substrate concentrations, enzyme activity increases steeply as substrate concentration increases. Random collisions between substrate and active site happen more frequently with higher substrate concentrations. At high substrate concentrations, most of the active sites are occupied, so raising the substrate concentration has little effect on enzyme activity and the graph begins to plateau.
3. Compare competitive/non-competitive inhibition.
Competitive inhibition is when an inhibitor similar to the substrate binds to the active site, blocking the active site and preventing substrate binding. The presence of a competitive inhibitor reduces the rate of reaction, but increasing the concentration of substrate can reduce the effect of the inhibitor.
Non-competitive inhibition is when an inhibitor not similar to the substrate binds to a site other than the active site (allosteric site) and changes the active site, preventing the substrate from binding to the active site.
4. Discuss lock-and-key vs. induced-fit.
The lock-and-key hypothesis states that enzymes are substrate-specific and the substrate fitting in the enzyme is like a key fitting in a lock. This does not explain the broad specificity of some enzymes. The induced-fit model helps explain how substrates change to fit an enzyme. The induced-fit model says the active site may undergo conformational change to better fit the substrate. The active site may change shape or configuration. This change is induced by the substrate.
Denaturation is changing the structure of an enzyme (or other protein) so that it can no longer carry out its function. Enzymes are denatured at very high temperatures and extreme pHs.
2.What does a graph for increasing substrate concentration of an enzyme look like? Describe it...
At low substrate concentrations, enzyme activity increases steeply as substrate concentration increases. Random collisions between substrate and active site happen more frequently with higher substrate concentrations. At high substrate concentrations, most of the active sites are occupied, so raising the substrate concentration has little effect on enzyme activity and the graph begins to plateau.
3. Compare competitive/non-competitive inhibition.
Competitive inhibition is when an inhibitor similar to the substrate binds to the active site, blocking the active site and preventing substrate binding. The presence of a competitive inhibitor reduces the rate of reaction, but increasing the concentration of substrate can reduce the effect of the inhibitor.
Non-competitive inhibition is when an inhibitor not similar to the substrate binds to a site other than the active site (allosteric site) and changes the active site, preventing the substrate from binding to the active site.
4. Discuss lock-and-key vs. induced-fit.
The lock-and-key hypothesis states that enzymes are substrate-specific and the substrate fitting in the enzyme is like a key fitting in a lock. This does not explain the broad specificity of some enzymes. The induced-fit model helps explain how substrates change to fit an enzyme. The induced-fit model says the active site may undergo conformational change to better fit the substrate. The active site may change shape or configuration. This change is induced by the substrate.
Friday, October 4, 2013
Warm Up - 10/4
Explain the process of aerobic cellular respiration.
Aerobic cellular respiration begins with glycolysis, the conversion of a molecule of glucose into two pyruvate molecules and two ATP molecules. The first step of glycolysis is phosphorylation, where hexokinase and then phosphofructokinase add two phosphates to the glucose molecule, creating a hexose biphosphate. The second step of glycolysis is lysis, the splitting of hexose biphosphate into two 3-carbon sugars by the enzyme aldolase. These 3-carbon sugars are called G3P, PGAL, or triose phosphate. The third step of glycolysis is oxidation where each G3P molecule loses two atoms of hydrogen by converting NAD+ into NADH + H+. The creation of NADH releases energy to add a phosphate group to both G3Ps. The resulting 3-carbon molecules with extra phosphate are called PGA. In the fourth step of glycolysis, ATP formation, ADP undergoes substrate-level phosphorylation, producing ATP. The remaining carbon and phosphate molecule is turned into pyruvate.
Fatty acids can also be used to bypass glycolysis. Since fatty acids are a long chain of carbon molecules, coenzyme CoA oxidizes this chain and breaks it down, making Acetyl CoA with two carbons.
After glycolysis, the link reaction occurs. Pyruvate is decarboxylated to form an acetyl group (2C) and CO2 is released. Then the pyruvate is oxidized and NAD+ accepts a hydrogen to form NADH + H+. The acetyl group combines with a coenzyme (CoA) to form Acetyl CoA, which enters the Krebs cycle.
In the Krebs Cycle, Acetyl CoA combines the acetyl group (2C) with oxaloacetate (4C), forming citrate (6C). Citrate undergoes oxidative decarboxylation, converting NAD+ into NADH + H+ and producing CO2. This forms a 5 carbon molecule, which undergoes oxidative decarboxylation again, making a 4 carbon molecule. Substrate level phosphorylation occurs, adding a phosphate to ADP to create ATP. FAD2 is converted into FADH2 and NAD+ is converted into NADH + H+. After these reactions, the 4 carbon molecule is converted into oxaloacetate, which is used in the next turn of the Krebs Cycle. The products of the Krebs Cycle are used to power the Electron Transport Chain.
Aerobic cellular respiration begins with glycolysis, the conversion of a molecule of glucose into two pyruvate molecules and two ATP molecules. The first step of glycolysis is phosphorylation, where hexokinase and then phosphofructokinase add two phosphates to the glucose molecule, creating a hexose biphosphate. The second step of glycolysis is lysis, the splitting of hexose biphosphate into two 3-carbon sugars by the enzyme aldolase. These 3-carbon sugars are called G3P, PGAL, or triose phosphate. The third step of glycolysis is oxidation where each G3P molecule loses two atoms of hydrogen by converting NAD+ into NADH + H+. The creation of NADH releases energy to add a phosphate group to both G3Ps. The resulting 3-carbon molecules with extra phosphate are called PGA. In the fourth step of glycolysis, ATP formation, ADP undergoes substrate-level phosphorylation, producing ATP. The remaining carbon and phosphate molecule is turned into pyruvate.
Fatty acids can also be used to bypass glycolysis. Since fatty acids are a long chain of carbon molecules, coenzyme CoA oxidizes this chain and breaks it down, making Acetyl CoA with two carbons.
After glycolysis, the link reaction occurs. Pyruvate is decarboxylated to form an acetyl group (2C) and CO2 is released. Then the pyruvate is oxidized and NAD+ accepts a hydrogen to form NADH + H+. The acetyl group combines with a coenzyme (CoA) to form Acetyl CoA, which enters the Krebs cycle.
In the Krebs Cycle, Acetyl CoA combines the acetyl group (2C) with oxaloacetate (4C), forming citrate (6C). Citrate undergoes oxidative decarboxylation, converting NAD+ into NADH + H+ and producing CO2. This forms a 5 carbon molecule, which undergoes oxidative decarboxylation again, making a 4 carbon molecule. Substrate level phosphorylation occurs, adding a phosphate to ADP to create ATP. FAD2 is converted into FADH2 and NAD+ is converted into NADH + H+. After these reactions, the 4 carbon molecule is converted into oxaloacetate, which is used in the next turn of the Krebs Cycle. The products of the Krebs Cycle are used to power the Electron Transport Chain.
Wednesday, October 2, 2013
Warm Up 2 - 10/2/13
1(a)(i) Mitochondria
(ii) The structure labelled I above is the cristae. Cristae folds are shelf-like projections of the inner membrane which increase the surface area available for oxidative phosphorylation.
(ii) The structure labelled I above is the cristae. Cristae folds are shelf-like projections of the inner membrane which increase the surface area available for oxidative phosphorylation.
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